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The Inverse Galois Problem: the Rigidity Method

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The Inverse Galois Problem: the Rigidity Method The Inverse Galois Problem: The Rigidity Method Amin Saied Department of Mathematics, Imperial College London, SW7 2AZ, UK CID:00508639 June 24, 2011 Supervisor: Professor Martin Liebeck Abstract The Inverse Galois Problem asks which finite groups occur as Galois groups of extensions of Q, and is still an open problem. This project explores rigidity, a powerful method used to show that a given group G occurs as a Galois group over Q. It presents the relevant theory required to understand this approach, including Riemann's Existence Theorem and Hilbert's Irreducibility Theorem, and applies them to obtain specific conditions on the group G in question, which, if satisfied say that G occurs as a Galois group over the rationals. These ideas are then applied to a number of groups. Amongst other results it is explicitly shown that Sn;An and P SL2(q) for q 6≡ ±1 mod 24 occur as Galois groups over Q. The project culminates in a final example, highlighting the power of the rigidity method, as the Monster group is realised as a Galois group over Q. Contents 1 Introduction 2 1.1 Preliminary Results . 2 1.2 A Motivating Example . 4 2 Riemann's Existence Theorem 5 2.1 The Fundamental group . 5 2.2 Galois Coverings . 7 2.3 Coverings of the Punctured Sphere . 8 3 Hilbertian Fields 13 3.1 Regular Extensions . 13 3.2 Hilbertian Fields . 14 3.3 Galois Groups Over Hilbertian Fields . 19 3.3.1 Sn as a Galois group over k .......................... 20 4 Rigidity 21 4.1 Laurent Series Fields . 21 4.2 Finite Extensions of Λ . 22 4.3 Branch Points . 24 4.4 Rigid Ramification Types . 26 4.5 Structure Constants . 28 5 The Rigidity Criteria 33 5.1 Descent of the Base Field . 33 5.2 Rigidity Criteria . 38 6 Applications of Rigidity 41 6.1 Sn and An as Galois groups over Q .......................... 41 6.2 P SL2(q) for q 6≡ ±1 mod 24 as a Galois group over Q . 43 6.3 M12 as Galois group over Q .............................. 46 2 2 ab 6.4 The Ree Groups G2(q ) as Galois groups over Q . 47 6.5 Realising the Monster over Q ............................. 49 7 Conclusion 51 References 52 A GAP Code 53 B Characters for M12 55 1 Chapter 1 Introduction By associating a certain group to a given polynomial, Evariste´ Galois was able to prove that there is no general solution by radicals to a polynomial equation of degree 5 or higher. His approach provided not only a beautiful answer to this classical problem, but also offered a great insight into why it is possible to solve polynomial equations of degree less than 5 in general. Galois' idea was to associate to a polynomial p(x) a certain group Gal(p(x)) which permutes the roots of p(x). In general, given a polynomial with rational coefficients, the Galois group is defined on a certain field extension E=Q, where E is the splitting field of p(x). It is here that a natural question arises. Inverse Galois Problem: Which groups can be realised as Galois groups over Q? The problem was first approached by Hilbert, in 1892, and remains unsolved today. This project develops the most successful approach to the question thus far, namely the concept of rigidity. Introduced by John Thompson in 1984 in his breakthrough paper [11], this method has already been used to realise numerous families of groups including, and perhaps most notably, all but 2 of the 26 sporadic groups. A striking feature of the material is the vast diversity of topics it draws from, including group theory, field theory, Galois theory, algebraic topology, Riemann surface theory and number the- ory. This project develops the relevant theorems from these areas and pulls them together to form a strict group theoretic condition, rigidity, which if satisfied in a group G, along with some other conditions, will guarantee a positive solution to the Inverse Galois Problem for G. Some background material is set up and some basic results from Galois theory are stated. Proofs, where not provided, and a more thorough description of the background can be found in [2] and [10]. 1.1 Preliminary Results A field extension E=K is algebraic if every element of E is algebraic over F . An algebraic field extension is said to be normal if every irreducible polynomial f(x) 2 K[x] which has a root in L, has all its roots in L. An algebraic field extension E=K is separable if every x 2 L has a separable minimal polynomial over K, that is, the minimal polynomial has distinct roots. A Galois extension is an algebraic field extension E=K which is normal and separable. The degree of an extension E=K is the dimension of E as a vector space over K and is denoted 2 jE : Kj. An extension is said to be finite if it has finite degree. Moreover for nested extensions E=K, K=F , the degree of E=F obeys the tower law jE : F j = jE : KjjK : F j as in [10]. Definition 1.1.1. Let E=K be a finite Galois extension. Then the Galois group is defined as Gal(E=K) := fσ 2 Aut(E): σ(x) = x 8 x 2 Kg This next basic result is fundamental to Galois theory and will be used throughout, often without reference. Lemma 1.1.2. Let f(x) 2 K[x] and let E=K be a Galois extension. If α 2 E is a root of f(x) and σ 2 Gal(E=K), then σ(α) 2 E is a root of f(x). n n Proof. Let f(x) = anx + ··· + a0 with ai 2 K. Now f(α) = 0 so anα + ··· + a0 = 0. Now applying σ 2 Gal(E=K) gives n n σ(anα + ··· + a0) = anσ(α) + ··· + an = 0 because σ fixes K pointwise. Hence f(σ(α)) = 0 Theorem 1.1.3. (The Primitive Element Theorem) Let E=K be a finite Galois extension. Then 9 α 2 E such that E = K[α]. I now quote the main theorem of Galois theory, from [2]. Theorem 1.1.4. Let E=K be a finite Galois extension with Galois group G = Gal(E=K). For an intermediate field F such that K ⊆ F ⊆ E let λ(F ) be the subgroup of G leaving F fixed. Then λ is a one-to-one map of the set of all such intermediate fields F onto the set of subgroups of G with the following properties: 1. λ(F ) = Gal(E=F ) 2. F = EGalE=F = Eλ(F ) where EH denotes the fixed field of H ≤ G in E. 3. For H ≤ G, λ(EH ) = H 4. jE : F j = jλ(F )j, jF : Kj = jG : λF )j 5. F is a normal extension of K if and only if λ(F ) is a normal subgroup of G. When this is the case . Gal(F=K) = Gal(E=K) Gal(E=F ) This correspondence between subgroups of Gal(E=K) and intermediate fields F between E and K is referred to as the Galois correspondence, and it too will be used throughout this project. I now quote another classic result from [10]. Theorem 1.1.5. (Artin's Theorem) If G is a finite group of automorphisms of a field E and if the fixed field of G in E is K then E=K is a finite Galois extension with Gal(E=K) = G 3 1.2 A Motivating Example How does one find a group as a Galois group? In the true nature of the original idea of Galois theory, one might think to look at polynomials f(x) 2 Q[x]. To find the Galois group of a poly- nomial one must find its splitting field Ef , which is the smallest field extension of Q containing the roots of f [10]. Lets consider the dihedral group of the square D8. Can it be realised as a Galois group over Q? Consider the irreducible polynomial 4 2 f(x) = x − 6x + 2 2 Q[x] In general the splitting field Ef of an irreducible polynomial of degree n can have degree up to n! over Q, and thus the Galois group is a subgroup of Sn [10]. Therefore in our case the splitting field Ef has degree dividing 24 (by Lagrange). This is why it was a good idea to look at a polynomial of degree 4, as D8 is the unique subgroup of S4 of order 8. Proceed by findingp 2 2 the roots of f(x) by first setting y = x and finding thep roots of y − 6y + 2, which are 3 ± 7. An extension of degree 2 has been created, namely Q( 7). Now find the roots of f(x) to be q p q p ±α := 3 + 7 ± β := 3 − 7 p p Hence the splitting field of f is E = (α; β). Notice that αβ = 9 − 7 = 2 2= (α). Now f Q p Q p 2 2 p β 2 Q(α) () αβ 2 Q(α). Therefore β2 = Q(α). Each equation x = 3+ 7 and x = 3 − 7 produces an extension of degree 2, and so we have that Ef is a degree 8 extension of Q. Therefore jGal(f)j = jGal(Ef =Q)j = 8, and since D8 is the unique subgroup of S4 of order 8 we have that 4 2 ∼ Gal(x − 6x + 2) = D8 So by a clever choice of polynomial one was able to realise D8 as a Galois group over Q.
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